Answer
Hyperbola:
Foci: $(\pm 2\sqrt 5, 0)$
Vertices: $(\pm 2, 0)$
Asymptotes: $y=\pm2x$
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Work Step by Step
1. Divide both sides of the equation by 16:
$$\frac{x^2}{4} - \frac{y^2}{16} = 1$$
According to the pattern, $a^2$ = 4 and $b^2$ = 16
$c = \sqrt {a^2 + b^2} = \sqrt {4 + 16} = \sqrt {20} = 2\sqrt 5$
$a = \sqrt {4} = 2$
$b = \sqrt {16} = 4$
And since "a" is under "$x^2$", the foci and vertices are determined by $(\pm c, 0)$ and $(\pm a, 0)$, respectively.
Foci = $(\pm 2\sqrt 5, 0)$
Vertices = $(\pm 2, 0)$
2. Sketch the graph
Use the vertices and the foci to sketch the graph of that equation: $4x^2 + y^2 = 16$
